MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus, Fall 2007 Transcript – Lecture 29 The following content is provided under a Creative Commons license. Your support will help MIT OpenCourseWare continue to offer high quality educational resources for free. To make a donation or to view additional materials from hundreds of MIT courses, visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: Today we're going to continue our discussion of methods of integration. The method that I'm going to describe today handles a whole class of functions of the following form. You take P (x) / Q (x) and this is known as a rational function. And all that means is that it's a ratio off two polynomials, which are these functions P ( x) and Q ( x). We'll handle all such functions by a method which is known as partial fractions. And what this does is, it splits P / Q into what you could call easier pieces. So that's going to be some kind of algebra. And that's what we're going to spend most of our time doing today. I'll start with an example. And all of my examples will be illustrating more general methods. The example is to integrate the function 1 / x -1 +, say, 3 / x + 2 dx. That's easy to do. It's just, we already know the answer. It's ln x - 1 + 3 ln x + 3. Plus a constant. So that's done. So now, here's the difficulty that is going to arise. The difficulty is that I can start with this function, which is perfectly manageable. And than I can add these two functions together. The way I add fractions. So that's getting a common denominator. And so that gives me x + 2 here + 3 ( x - 1). And now if I combine together all of these terms, then altogether I have 4x + 2 - 3, that's - 1. And if I multiply out the denominator that's x ^2 + that 2 turned into a 3, that's interesting. Hope there aren't too many more of those transformations. Is there another one here? STUDENT: [INAUDIBLE] PROFESSOR: Oh, it happened earlier on. Wow that's an interesting vibration there. OK. Thank you. So, I guess my 3's were speaking to my 2's. Somewhere in my past. OK, anyway, I think this is now correct. So the problem is the following. This is the problem with this. This integral was easy. I'm calling it easy, we already know how to do it. Over here. But now over here, it's disguised. It's the same function, but it's no longer clear how to integrate it. If you're faced with this one, you say what am I supposed to do. And we have to get around that difficulty. And so what we're going to do is we're going to unwind this disguise. So we have the algebra problem that we have. Oh, wow. There must be something in the water. Impressive. Wow. OK, let's

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